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Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6506

Exercise

Given the differentiable function

z=\frac{y^2}{3x}+f(xy)

Prove the equation

x^2z'_x-xyz'_y+y^2=0

Proof

Define

u=xy

We get the function

z=\frac{y^2}{3x}+f(u)

And the internal function

u(x,y)=xy

We will use the chain rule to calculate the partial derivatives of z.

z'_x=\frac{y^2}{3}\cdot (-\frac{1}{x^2})+f'_u\cdot u'_x=

=-\frac{y^2}{3x^2}+f'_u\cdot y

 

z'_y=\frac{1}{3x}\cdot 2y+f'_u\cdot u'_y=

=\frac{2y}{3x}+f'_u\cdot x

We will put the partial derivatives in the left side of the equation we need to prove.

x^2z'_x-xyz'_y+y^2=

=x^2(-\frac{y^2}{3x^2}+yf'_u)-xy(\frac{2y}{3x}+xf'_u)+y^2=

=-\frac{y^2}{3}+x^2yf'_u-\frac{2y^2}{3}-x^2yf'_u+y^2=

=-\frac{y^2}{3}-\frac{2y^2}{3}+y^2=

=y^2(-\frac{1}{3}-\frac{2}{3}+1)=

=y^2\cdot 0

= 0

We got zero as required.

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